Why Are Inverses Unique in a Category?

[Math Amateur]

I am reading Steve Awodey's book: Category Theory (Second Edition) and am focused on Section 1.5 Isomorphisms ...

I need some further help in order to fully understand some further aspects of Definition 1.3, Page 12, including some remarks Awodey makes after the text of the definition ... ...

The start of Section 1.5, including Definition 1.3 ... reads as follows:View attachment 8354In the text of Definition 1.3 we read the following:

" ... ... Since inverses are unique (proof!), we write \(\displaystyle g = f^{-1}\). ... ... "Can someone please demonstrate a rigorous proof that in a category, inverses are unique ... ?Help will be appreciated ...

Peter

 

[steenis]

You have a isomorphism $f:A \rightarrow B$ in a category $C$

Suppose, there are two arrows $g,h:B \rightarrow A$ such that

$fg=1_B$, $gf=1_A$, $fh=1_B$, and $hf=1_A$

in other word, each is an inverse of $f$, we have to prove that $g=h$

Consider $hfg:B \rightarrow A \rightarrow B \rightarrow A$

On one hand, we have $hfg=h(fg)=h1_B=h$

On the other hand, we have $hfg= \cdots$, can you finish this ?

 

[Math Amateur]

You have a isomorphism $f:A \rightarrow B$ in a category $C$

Suppose, there are two arrows $g,h:B \rightarrow A$ such that

$fg=1_B$, $gf=1_A$, $fh=1_B$, and $hf=1_A$

in other word, each is an inverse of $f$, we have to prove that $g=h$

Consider $hfg:B \rightarrow A\rightarrow B$

On one hand, we have $hfg=h(fg)=h1_B=h$

On the other hand, we have $hfg= \cdots$, can you finish this ?

Thanks Steenis ...

Hmm ... easy when you see how ... :) ...

We have ...

\(\displaystyle hfg = h(fg) = h 1_B = h\) ... ... ... ... ... (1)

and

\(\displaystyle hfg = (hf)g = 1_A g = g\) ... ... ... ... ... (2) ... so it follows that ...

(1) (2) \(\displaystyle \Longrightarrow g = h\)Hope that is correct ...Thanks again Steenis ...

Peter

 

[steenis]

Yes that is correct

You did these things before, for instance, in Group Theory

In post #2, I meant $hfg:B \rightarrow A \rightarrow B \rightarrow A$
Already corrected